general indefinite integral?

Question

sinx / (1-sin^2x) dx
I'm not sure what to do with the problem, I've got the top atleast!

Answer 1

The bottom part is equal to cos^2 x.
(Follows from fundamental relation sin^2 + cos^2 = 1).

Now you have

sin x / cos^2 x dx

At the bottom a function, at the top its derivative ... that is beautiful. We can write sin x = -d(cos x), so we must find integral of

-d(cos x) / (cos x)^2

or, with u = cos x, of

-u^(-2) du

The indef. integral of this is u^(-1) or 1/u, so the solution is

1/cos x + C



P.S. @ Goose... the integral of u^(-2) is -u^(-1), no factor two involved.

Answer 2

easy if sin^2x+cos^2x = 1 then 1-sin^2x = cos^2x

so sinx/cos^2x dx is your new integral. Set cos(x) = u so du = sin(x)dx so you end up with du/u^2.

so du/u^2 is the same as u^-2du which comes out to.... -2/u but cos(x) = u so it is -2/cos(x) so all said and done is:

-2sec(x)

Answer 3

1 / cos(x)